Bowling balls rolling up a ramp (conservation of energy)

[macaholic]

This is from an old course I took. I'm not sure what I'm doing incorrectly.

Homework Statement

Two identical bowling balls are rolling on a horizontal floor without slipping. The initial speed of both balls is V = 9.9 m/s. Ball a encounters a frictionless ramp, reaching a maximum height H_a. Ball B rolls up a regular ramp (i.e. rolls without slipping), reaching a maximum vertical height H_b. Which ball goes higher, and by how much?

Homework Equations

$GPE = mgh$
$KE_{trans} = \frac{mv^2}{2}$
$KE_{rot} = \frac{I\omega ^2}{2}$
$I_{solid-sphere} = \frac{2 M R^2}{5}$
$\omega = \frac{v}{R}$

The Attempt at a Solution

I just did conservation of energy. Ball a will still have rotational energy when it reaches the top, ball b will not. Both balls start with the same total kinetic energy. Ball a's final GPE will come entirely from its initial translation energy. Ball b's will come from both its rotational and translational kinetic energies. Solving:
$\frac{mv^2}{2} = mgH_a$
$\frac{mv^2}{2} (1 + \frac{1}{5}) = mgH_b$
$H_a = \frac{v^2}{2g}$
$H_b = \frac{3v^2}{5g}$

However, the answer says that I should get that H_b is "2m" higher than H_a. Did I do something wrong here?

 

[MostlyHarmless]

I think you should be thinking about it like this: Ball A and B have some total energy, both equal to ${\frac{1}{2}}mv^2$. The must have the same amount of energy at any point during the motion. Ball A's total energy will be converted entirely into potential energy. Which you have done. Ball B's total energy will be converted into rotational kinectic energy and gravitational potential energy so ${\frac{1}{2}}mv^2=mgH_b+{\frac{1}{5}}mv^2$

I'm not very familiar with the rotational kinectic energy, so I'm assuming you are right about it. I'm commenting on the setup of your equations.

 

[TSny]

$\frac{mv^2}{2} (1 + \frac{1}{5}) = mgH_b$

Check to make sure the fraction 1/5 is correct here.

 

[macaholic]

Except that both balls don't always have total energy equal to $\frac{mv^2}{2}$. They're said to be rolling withoutt slipping initially, which means that they have both translattional and rotational kinetic energy, i.e. $KE_{tot} = \frac{mv^2}{2} + \frac{I\omega^2}{2}$.

Ball a is still rolling with when it reaches its highest point, hence its full conservation of energy equation should look like this:
$KE_{trans} + KE_{rot} = GPE + KE_{rot}$
$\frac{mv^2}{2} + \frac{I\omega^2}{2} = mgH_a + \frac{I\omega^2}{2}$

That's where my equations come from. I was fairly certain that part was okay but I could be wrong...

 

[rcgldr]

$KE_{rot} = \frac{I\omega ^2}{2}$
$I_{solid-sphere} = \frac{2 M R^2}{5}$
$\omega = \frac{v}{R}$

If you follow your equations from above, you have:

$KE_{b} = \frac{mv^2}{2} + \frac{mv^2}{5}$

but then you factor out $\frac{mv^2}{2}$ from $\frac{mv^2}{5}$ and get 1/5?

 

[macaholic]

@Tsny, I will try to be more explicit, let's see...

For B:
$KE_{trans} + KE_{rot} = GPE$
$\frac{mv^2}{2} + \frac{1}{2} \frac{2mR^2\omega^2}{5} = mgH_b$
$v^2(\frac{1}{2} + \frac{1}{5}) = gH_b$
$H_b = \frac{7v^2}{10g}$

 

[voko]

$\frac{mv^2}{2} (1 + \frac{1}{5}) = mgH_b$

Where does 1/5 come from?

$H_b = \frac{3v^2}{5g}$

Why 3/5?

However, the answer says that I should get that H_b is "2m" higher than H_a.

What is 'm'? The unit meaning meters?

 

[TSny]

@Tsny, I will try to be more explicit, let's see...

For B:
$KE_{trans} + KE_{rot} = GPE$
$\frac{mv^2}{2} + \frac{1}{2} \frac{2mR^2\omega^2}{5} = mgH_b$
$v^2(\frac{1}{2} + \frac{1}{5}) = gH_b$
$H_b = \frac{7v^2}{10g}$

Looks good.

 

[macaholic]

Thanks guys, I stink at algebra apparently haha. So is this stuff about the "2m" nonsense problaby just some sort of weird mistake?

 

[rcgldr]

$H_b = \frac{7v^2}{10g}$

That fixes it, 7/10 instead of 3/5.

 

[voko]

Thanks guys, I stink at algebra apparently haha. So is this stuff about the "2m" nonsense problaby just some sort of weird mistake?

If 'm' is 'meters', then it makes perfect sense. What number do you get?

 

[TSny]

Thanks guys, I stink at algebra apparently haha. So is this stuff about the "2m" nonsense problaby just some sort of weird mistake?

I think the "2m" means 2 meters and not 2 times the mass!

 

[macaholic]

What is 'm'? The unit meaning meters?

No idea. It just says "2m higher". I didn't actually plug in the numbers though...

H_a = 5 meters
H_b = 7 meters.

WELP. Mystery solved. It's listed in italics... so I didn't think it was a unit. Thanks again haha.

 

[macaholic]

Thanks for the tag team help everyone :). I feel slightly silly now.

 

[SammyS]

No idea. It just says "2m higher". I didn't actually plug in the numbers though...

H_a = 5 meters
H_b = 7 meters.

WELP. Mystery solved. It's listed in italics... so I didn't think it was a unit. Thanks again haha.

Yup.

(9.9 m/s )² = 98.01 m²/s²,

so v²/g is very close to 10 meters.